In dajmcdon/ubc-stat406-labs: Tutorials and labs for UBC Stat 406 in the 2020-2021 online year

Recall "Maximum likelihood estimation". We're going to do this by hand. The Poisson Likelihood is given by [ L(\lambda ; y_1,\ldots,y_n) = \prod_{i=1}^n \frac{\lambda^{y_i}\exp(-\lambda)}{y_i!}. ]

1. Generate $n=100$ observations from a Poisson distribution with parameter $\lambda=3$.

y = rpois(100, 3)

1. Examine the slide from class. What key ingredient do I need to make this work?

I need the derivative of the likelihood.

1. Remember that if I use a monotone function like log, I don't change the minimizer. Use this trick!! Write a function which gives the key ingredient as a function of $\lambda$.

deriv_loglike <- function(lambda, y){
  sum(-1 + y/lambda)
}

1. Suppose $\lambda=1$, $y_1=3$, $y_2=5$. What value should your function return? Check that it does.

You should get $3+5-2=6$. My function gives r deriv_loglike(1,c(3,5)).

1. Using your data and my sample code from class, maximize the loglikelihood and report the maximizer. Use gam=.01 and start at 23. Set tol=1e-5. Is this the value you would expect? What should the answer be?

I should see $\lambda = \overline{y} =$ r mean(y).

maxiter = 1000
conv = FALSE
gam = 0.01
lam = 23
tol = 1e-5
for(iter in  1:maxiter){
  lam.new = lam + gam * deriv_loglike(lam, y)
  conv = abs(lam - lam.new) < tol
  lam = lam.new
  if(conv) break
}
lam

dajmcdon/ubc-stat406-labs documentation built on Aug. 18, 2020, 1:23 p.m.